1
$\begingroup$

I and my opponent play a game of dices. I win if my opponent didn't throw a dice in our game with a higher number than my highest number.

Example : I have 2 dices, he's got 3 . Our throws : (1,4,2,4,3) , I threw 1,4 he 2,3,4. I won in this case because he didn't get any higher number than my 4. (he should've gotten a 5 or 6) The dices are also distinguishable.

Give the number of possible combinations in which I can win if:

1) I have 2 dices, he's got 1.

2) I have 3 dices and my opponent 2.

My incomplete solution:(let $i$ be my throws and $o$ for opponent)

1) (i1, i2, o1) here are possible $6^3$ total combinations .

$wins=6^3 - |o1>max(i1,i2)|$ (but I don't know how to count it)

2) no idea

Thank you.

$\endgroup$
  • $\begingroup$ Are you expected to give the number of possible combinations, or list out all the combinations? There are quite a few of them for part (2). $\endgroup$ – Brian Tung Apr 21 '17 at 17:44
  • $\begingroup$ @BrianTung count all possible combinations in which I turn out as a winner. $\endgroup$ – Oleg Apr 21 '17 at 17:44
  • $\begingroup$ Are the dice distinguishable? In other words, in (1), if you roll a 4 and a 5, is that different from rolling a 5 and a 4? $\endgroup$ – Brian Tung Apr 21 '17 at 17:46
  • $\begingroup$ These numbers are small enough that I would just start listing the possibilities out. You may begin to see some regularities that will permit you to count them more efficiently. Since the dice are not distinguishable, there are not, in fact, $6^3 = 216$ combinations of three dice. (For example, 1/2/3 is not different from 3/2/1.) This complicates the count quite a bit. $\endgroup$ – Brian Tung Apr 21 '17 at 17:51
  • 1
    $\begingroup$ You’re comparing the maximum of your die rolls to the minimum of your opponent’s, so you can use order statistics to do this calculation. $\endgroup$ – amd Apr 21 '17 at 18:05
2
$\begingroup$

In general, if you have $\max(Z_1, Z_2, Z_3, ..., Z_n) = m$ for dice $Z_i$ and max outcome $m$, then you have $M_{m,n} = \sum_{k=1}^{n} \binom{n}{k} (m-1)^{n-k}$ ways to get that outcome.

You choose $k$ of the dice to equal $m$, and then the remaining $n-k$ dice can take on one of $m-1$ remaining values each. This simplifies to $M_{m,n} = m^n - (m-1)^n$

So for each max we can achieve, $m_1$, we wish to compute the number of ways we can achieve that max, multiplied by the number of ways the opponent achieves a max $m_2 \leq m_1$.

Let $m_1, a$ be our max and number of dice, and let $m_2, b$ be the opponent's max and number of dice. Let $F(a, b)$ be the number of ways we can win given $a, b$ dice.

$$F(a, b) = \sum_{m_1=1}^{6} \sum_{m_2=1}^{m_1} (m_1^{a} - (m_1-1)^{a})(m_2^{b} - (m_2-1)^{b})$$

This can be simplified to a closed-form expression:

$$F(a, b) = 1 - 2^{b} + 2^{a + b} - 2^{a} 3^{b} + 3^{a + b} - 3^{a} 4^{b} + 4^{a + b} - 4^{a} 5^{b} + 5^{a + b} - 5^{a} 6^{b} + 6^{a + b}$$

See that $F(2, 1) = 161$ and $F(3, 2) = 5593$.

$\endgroup$
0
$\begingroup$

$(1).~161$
$(2).~1071$
combination are possible to win the game.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.